A spherical iron ball of radius 10cm is coated with a layer of ice of uniform thickness that melts a rate of $50 c m^{3} / \min$ . When the thickness that of the ice 5cm, then the rate at which the thickness in cm/min) of the ice decreases, is

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$\frac{5}{6 π}$

$\frac{1}{18 π}$

$\frac{1}{9 π}$

$\frac{1}{36 π}$

answer is B.

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Detailed Solution

$\frac{d v}{d t} = 50 c m^{3} / \min$
Thick news of ice r = 5cm
$V = \frac{4}{3} π r^{3}$
$\frac{d v}{d t} = \frac{4}{3} π (3 r^{2}) \frac{d v}{d t}$
$\Rightarrow \frac{d v}{d t} = \frac{1}{18 π} c m / \min$

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